03 Planes and linear spaces, part 1

Planes in \(\mathbb{R}^3\)

Point-slope equations of planes

Point-slope equation for a line in \(\mathbb{R}^2\)

• The point-slope equation for a line through a point \((x_0,y_0)\) with slope \(m\) is \[ y = m(x-x_0) + y_0. \]

• We interpret the slope \(m\) as the slope in the positive \(x\)-direction.

Point-slope equation for a plane in \(\mathbb{R}^3\)

• The point-slope equation for a plane through a point \((x_0,y_0,z_0)\) with slope \(m\) in the positive \(x\)-direction and slope \(n\) in the positive \(y\)-direction is \[ z = m(x-x_0) + n(y-y_0) + z_0. \]

Exercise 1: Point-slope equations

1. Compute the point-slope equation for a plane through the point \((1,2,3)\) with slope \(2\) in the positive \(x\)-direction and slope \(-1\) in the positive \(y\)-direction.

2. The point-slope equation of a plane is \[ z = 4(x-2) - 3(y-2) + 5. \] Through what point does this plane pass? What are the slopes in the \(+x\) and \(+y\) directions?

3. The point-slope equation of a plane is \[ z = 3x+ 4. \] Through what point does this plane pass? What are the slopes in the \(+x\) and \(+y\) directions?

4. The point-slope equation of a plane is \[ z = 0. \] Through what point does this plane pass? What are the slopes in the \(+x\) and \(+y\) directions?

Exercise 2: Linear models

The profit, \(p\), that a company makes depends on the number of two types of gadgets that it produces, Gadget 1 and Gadget 2. Let \(g_1\) and \(g_2\) be the number of gadgets that are manufactured of each type, and suppose they sell for \(s_1\) and \(s_2\) dollars (per gadget), respectively. Suppose the company believes that the profit depends linearly on the number of gadgets produced.

Let \(p = f(g_1,g_2)\) be the profit function, and assume that if no gadgets are products, then \(p=0\).

1. What is the domain of \(f\)? What is its codomain?

2. Find a formula for the profit function \(p = f(g_1, g_2)\).

Vector equations of planes

• Suppose a point \((x,y,z)\) lies on a plane with equation \[ z = m(x-x_0) + n(y-y_0) + z_0. \]
• The two points \((x,y,z)\) and \((x_0,y_0,z_0)\) have position vectors \[ \mathbf{r} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \quad \text{and} \quad \mathbf{r}_0 = \begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix}. \]
• If the vector \(\mathbf{r}-\mathbf{r_0}\) has its tail moved to the point \((x_0,y_0,z_0)\), then it lies in the plane. Draw the picture!
• Define \[ \mathbf{n} = \begin{bmatrix} -m \\ -n \\ 1 \end{bmatrix}, \] which is an example of a normal vector to the plane. (See below.)
• Then, observe: \[ \langle \mathbf{n}, \mathbf{r}-\mathbf{r}_0 \rangle = \left\langle \begin{bmatrix} -m \\ -n \\ 1 \end{bmatrix}, \begin{bmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{bmatrix} \right\rangle = -m(x-x_0) -n(y-y_0) + (z-z_0) = 0. \]

Vector equation of a plane in \(\mathbb{R}^3\)

Let \(\mathbf{n}\) and \(\mathbf{r}_0\) be two given vectors in \(\mathbb{R}^3\). A vector equation of the plane with normal vector \(\mathbf{n}\), passing through the point with position vector \(\mathbf{r}_0\), is \[ \langle \mathbf{n}, \mathbf{r}-\mathbf{r}_0 \rangle = 0, \] where \(\mathbf{r}\) is a variable position vector in \(\mathbb{R}^3\).

• Any vector that is orthogonal to all vectors in a plane is called a normal vector to the plane.

Standard equations of planes

• If a plane has a point-slope equation, then a normal vector can always be found of the form \[ \mathbf{n} = \begin{bmatrix} -m \\ -n \\ 1 \end{bmatrix}, \] as we saw on the prevoius slide.
• But not all planes have normal vectors of this special form!
• So, we must resort to the vector equation. If we write \[ \mathbf{n} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}, \] and expand the inner product in the vector equation, we get the following:

Standard equation of a plane in \(\mathbb{R}^3\)

A standard equation of a plane in \(\mathbb{R}^3\) passing through the point \((x_0,y_0,z_0)\) is \[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0, \] where \(a\), \(b\), and \(c\) are components of a normal vector \(\mathbf{n}\).

Exercise 3: Normal vectors and standard equations

a. Practice with normal vectors

1. Find a normal vector to the plane \(z = 2(x-1) - (y-2) + 3\).
2. Find a normal vector to the plane \(z = 4(x-2) + 5\).
3. Find a normal vector to the plane \(z = -y\).
4. Find a normal vector to the plane \(x=2\).
5. Find a point-slope equation of the plane with normal vector \[ \mathbf{n} = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \] passing through the point \((1,2,3)\).

b. Normal vectors and the coordinate planes

1. Find a normal vector to the \(xy\)-plane in \(\mathbb{R}^3\).
2. Find a normal vector to the \(xz\)-plane in \(\mathbb{R}^3\).
3. Find a normal vector to the \(yz\)-plane in \(\mathbb{R}^3\).

b. Normal vectors and standard equations

1. Find a normal vector to the plane with standard equation \(2(x-2) - 3y + z = 0\). Identify a point on this plane.
2. A plane passes through the point \((1,2,-1)\) and has normal vector \(\mathbf{e}_2\). Find its standard equation.

Affine equations of planes

Affine equation of a plane in \(\mathbb{R}^3\)

An affine equation for a plane in \(\mathbb{R}^3\) is \[ ax + by + cz = d, \]

where:

• \(a\), \(b\), \(c\), and \(d\) are constants,
• not all of \(a\), \(b\), and \(c\) are \(0\), and
• the numbers \(a\), \(b\), and \(c\) are the components of a normal vector to the plane.

Exercise 4: Affine equations

a. From affine equations to standard equations

Convert the following affine equations to standard equations. Give a normal vector and a point on each plane.

1. \(2x - 3y + z = 5\)
2. \(x + y = 1\)
3. \(2y - z = 2\)
4. \(x = 4\)

b. From standard equations to affine equations

Convert the following standard equations to affine equations:

1. \(4(x-1) + 2(y-2) - 5(z-3) = 0\)
2. \((x-1) + 3(y-2) = 0\)
3. \(2(y-1) - (z-2) = 0\)
4. \(2(x-8) = 0\)

c. From affine equations to point-slope equations

Convert the following affine equations to point-slope equations, if possible.

1. \(2x - 3y + z = 5\)
2. \(6x + y = 1\)
3. \(2y - 3z = 2\)
4. \(x = 4\)